Séminaire Choquet.
Initiation à l’analyse
J ORAM L INDENSTRAUSS
Some open problems in Banach space theory
Séminaire Choquet. Initiation à l’analyse, tome 15 (1975-1976), exp. no 18, p. 1-9
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Seminaire CHOQUET
(Initiation
18-01
l’analyse)
anné e I- t975/76, nO’ 18 ,,
a
91 p~
8 avril 1976
SOME OPEN PROBLEMS IN BANACH SPACE THEORY
by Joram LINDENSTRAUSS
My aimnn this talk is to discuss briefly some directions of research;
morphic theory of Banach spaces. The’directions which I plan to discuss
following.
in the isoare
the
’
(a)
Existence of operators,
(b)
Local
(c)
The approximation
(d)
Existence of bases,
(e).
Non linear
(f)
Nbn
theory
of Banach spaces,
.
,
property ,
problems,
separable problems.
In each of these
illustrate the
directions, I shall discuss only very few open questions so as to
specific direction. It should be emphasized that the choice of these
questions
is based
damental)
open
on
my
questions
personal interest and there
in these directions which
many other
are
(often
more
fun-
currently being studied by
various mathematicians. All the topics mentioned above are concdrned
mainly with general Banach spaces. Because of lack of time, I shall not enter into another
very
active domain of research in the isomorplic theory of Banach
the
spaces..are
namely
structure of
special Banach spaces. Let me just mention that among the special Banach spaces the most widely studied ones are the classical Banach
spaces, i. e.
spaces
closely
related to
L (~)in
progress made in recent years
and
the
C(K)
’
study
spaces. In
spite of
the very
of these spaces, there
are
challenging open problems concerning them. Recently, there began also,
fort to study the structure of some non classical- spaces (like
spaces
(especially
functions
Cp
,1
of these
of
P
the disc
~ , of operators
spaces
still many
a
of
serious ef-
analytic
algebra),
spaces of operators (especially the spaces
and Orlicz spaces). The study of the structure
generates many interesting problems which
analysis like. harmonic
significant
are
related to other
analysis, probability, complex analysis
areas
and operator thee-
ry.
As far
as
references
latest results in the
to many earlier
Let
us
concerned, I shall give here mostly references to the
specific direction (these references contain in turn references
are
contributions).
start with
topic (a). The only general method for constructing operators.
between general Banach spaces X and Y is the Hahn-Banach theory This theorem
ensures the existence of continuous linear functionals on X
and thus of operators
1
of rank
from
X
Y.
to
By taking
get operators
of finite
rank, and by
taking limits, we get compact operators from X to Y’. At this stage, the general
construction methods stop. We shall come back a little bit later to the
question
what kind of compact operators do we actually get. First, I want to discuss the
question of existence of non compact operators. There are simple and well known
examples
we
sums,
of infinite dimensional Banash spaces X and Y such that
every bounded lineai
Y is
operator from X
Y
(take e. g. X
with p. > r ). In
order to get a meaningful question, we have to restrict the
An in-
to
compact
teresting situation occurs if X
operator namely the identity I .
precisely :
(Q.1~
Does there exist
ded linear operator
If such;
an
X
pairs (X, Y’)’ .
we always have a non compact
other non compact operators ? More
= Y. In this case,
Are there also
infinite dimensional Banach space X so that every bounX -~ X is of the form T
+ K
with K compact ?
an
T :
exists,
= ,~ p t
it would have. several other
interesting properties.
(i) X
is not isomorphic to its subspaces of finite-codimension
is either compact or a Fredholm operator of index
0 ).
(ii) X
X
=~ Y’ p
is
Z
indecomposable,
is trivial in the
(iii) Every
e.
sense
every decomposition of
that either dim:
bounded linear operator
T :
X -~- X
has
X
into
a
T : .X~ X
direct
sum
dim.
or
non
(every
trivial, closed invariant
subspaces.
There is
no
(concerning (iii),
properties
make it
non
example
of
riant
known example of
a
we
Banach
have
space which has either one of these three
to add the assumption that X is separable
trivial. It should be pointed out also that P. ENFLO
an
operator
on some
Banach space which fails to have
[7]
non
constructed
to
an
trivial inva-
subspaces).
Coming back to
to Y , we point
the
general available methods for constructing operators from X
out that the
the existence of operators
Such operators
were
T :
of the Hahn-Banach theorem insures in
X --. Y having the fonn
called by GROTHENDIECK
general only
nuclear operators. He posed the follo-
wing question.
(Q.2) Does
that
there exist
air of infinite dimensionsal Banach spaces X
ever compact operator from X
to Y is nuclear ?
a
This question is still open, but there
which seem to suggest that the answer to
negative if
convex
[5].
X
has
an
are
by
(Q.2)
unconditional basis [10],
now
several
if either
Y
so
strong partial results
is negative. The
or
and
X
answer
or
Y’
is
is
e.
g.
uniformly
question (Q.2)
The
is
actually already
a
question
(b)
in item
of the list above
the
theory of Banach spaces. This theory is devoted to the study
structure of finite-dimensional subspaces of Banach spaces or in other words to the
study of convex sets in the euclidean space l~n for n large: (but finite). If n
namely
the local
several phenomena which defy perhaps the common intuition
(derived from the cases n = 2 , 3 ). A central result in the local theory of Banach
0 and integer
spaces is the theorem of Dvoretzky which. asserts that for every e >
k there. is an integer n
n~s ~. k) so that every Banach space X of dimension
is large, there
occur
=
(in particular,
n
dim: X
if
=~)
has
dim.Y
with-
Y
subspace
a
=
k
and’
subspaces in an arbitrary
Banach spaces enables one to reduce some questions on general Banach spaces to the
much simpler case of Hilbert space. In general the almost hilbertian subspaces of
concrete finite-dimensional Banach spaces are of ~ surprisingly high dimension (cf.
The existence of finite-dimensional almost hilbertian
and this fact increases very much .the effectiveness ot the
[8],
bertian subspaces in the
useful if
study
could show that
one
use
of almost
general Banach spaces. They would be even more
there are "nicely situated". This leads to the folloof
wing question.
(Q.3) Let
exist
a
X
be
an
constant
C
infinite-dimensional unif orm~.
so
that f or every
n
(Yn , ln2) C and there is a projection
TZAFRIRI
stronger results
assumption that
X
=
that the
proved
C~~ , 1 ~
or
be
uniformly
L ~0 ~
X =
~;n .
complemented copies of
i;~
arbitrary Banach
(Q.4) Let
C
X
anda
spaces
Yn
p~:
~
be
an
(equal
X with
(Q.2);
infinite-dimensional Banach:
either to
so
d(Yn , lnp)
C
In the
study of
Nn and ln~
problem).
(Q.2)
a
has
the
.
does not have
nicely
nicely complemented copies of
th~t
X
(Q.3)
which makes
sense
a
(Q.4)
the "local
non
natural way
This is shown most
that f or every
are
Pn
from
a
constant
n
there is
X
on to
a
sub-
Yn with
containe.d in the above mentioned
point out that. positive answer to (Q.4) imnegative answer (this illustrates the claim made
the structure of
enter in
ta.
Does there exist
me
a
question in
space.
and aprojection
partial answers
that
is
C ?
unconditional bases. Some
an
version of
a
that
[91. Let us point out that
in (Q.3) since if e. g.
These, spaces have however
paper of JOHNSON and TZAFRIRI. Let
abov~ that
has
so
spaces.
The strongest known
plies immediately
X
easily verified
This observation. leads to
for
if
is essential
it is
X
with ~Pn~
Pn : X ~ Yn
positive
convex
Banach space. Does there
there is a subspace
be found in JOHNSON and TZAFRIRI
can
X
is
answer
convex
clearly
theory").
uniformly
(it
convex
Banach spaces the
is this fact whichimakes
(Q.4)
a
spades
reasonable
in the work. of R. C. JAMES. A result which
started
research direction in the local
a
to the direction initiated
Let
Y
X
of
be
X
of Banach spaces
theory
by Dvoretzkij’s theorem is
the
reflexive Banach space. Then for every e
that dim Y
2 and d(Y ,
t + e . For
=
known whether the
same
problem by proving
There is however
is true for
that the
answer
~
for every finite,
is negative for
n
=
>
0
a
long time, it
n .
3
parallel
following theorem
a non
so
which is
there is
of James.
subspace
a
was
not
JAMES settled this
(and
thus for all
n > 3 ).
stronger
reflexivity condition which implies the existence of
almost isometric copies of
This is obtained in the following way : To say that
X is not reflexive means that the canonical
is not onto :
isometry J : X
There are two canonical isometries from
into
and
a
non
~.
~ X**
X**
Either
both maps
(this
of these maps is onto if and
one
together yieldall
is the
only if
i.
if and only if
case:
this is not the
of
then
J **.
X
is reflexives One can ask whether
whether
a.
is
(Jx)**
namely
reflexive).
It
was
shown in
[6]
that if
X
contains almost isometric copies even of
the very many open problems in this direction let me mention
the following.
(Q.5) Let
for every
case
X
be
e
and every
Banach space such that
a
inteqer
n
We turn
is not reflexive 1~ it true that
there is
In particular, what is the situation for
Among
subspace
a
n
=
Y
X
of
with
.
5 ?
to the
now
approximation property. Let us recall that a Banach space has
approximation property (AP in short), if for every compact subset K of X and
every E > 0 there i s an operator T : X ~ X which
the
;_
for every
x e
K . ENFLO
the first who proved that there
was
are Banach spaces which
fail to have the approximation
property. Since Enflo’s work several other examples of
spaces which fail to have the AP were constructed. We
mention,in particular Davie~s
paper
[4]
which contains
elegant examples
fail to have
of
the AP and Szankowski’s example
to have- the AP. It is however still far
from
or
subspaces of lp , 2
[t6]
be
X
Hilbert space.
an
Does
infinite-dimensional Banach
X
have
a
a
being clear
fails to hold "in general". For
example., the
(Q.6). 1~
of
Banach lattice which fails
to what extent the AP holds
following problem
is open.
which is not isomorphic to
subspace which fails to have the AP ?
space
All known examples of spaces which fail
to have the AP are in some
cial". It is of interest to exhibit
spaces which appear naturally in
fail to have; the AP. This leads to the
following.
B(l2)
the operator
norm.
be
the
Does
space of all bounded operators from
B(~)
p . , which
l2
sense
a
"artifi-
analysis
and
into itself. wi th,
have the AP ?’ Does the space of all bounded
analytic
f(z)
functions
~z~ ~ ~
~z: ;
on
with the sup
Note that both spaces mentioned in
(Q.T)
norm.
have the AP ?’
separablee There are several
in separable spaces (the existence
are non
strong and simple criteria for checking the AP
of a Schauder basis for example), By rising such
criteria
a
simply by
or even more
arguing directly it is easy to prove. that the common separable spaces which appear
in analysis have the AP. There is a variant of the AP called the uniform approxima~
(UAP
tion property
blems in the
in
short)
which leads to
of several
setting
interring
and
perhaps difficult
pro-
separable spaces. This property was introduced by PELCZYNSKI and ROSENTHAL and the point in it is that it does not only ask
whether
suitable
a
dim TX
on
n ~
T : X
a
the
on
X
->
X -> X
with
A Banach space
co .
f(n)
T :
common
integers
exists but asks for
is said to have the
that for every
so
with
X
dim. TX
03BB , Txi =
1
for
xi
i
Banach space is said to have the UAP if it has the
CZYNSKI and ROSENTHAL proved that the spaces L
however
use.
of
a
that
sely
basis in
X
even
(this
the existence of
follows from
C(K), and
spaces ~12~~
related.. to
Does the space
L
P
(Qc~)
Problems
that
and
(Q. $ )
spaces)
( the
C(l2)
norm) have the UAP ?
operator
[12]
and
has the UAP
xive Orliez
(Q.8,)
X
Does the
an
~y ~~ ) ~
which
space of
n , and
in
dim
X
C(K)
and
estimate
a
function
there is
f(n) .
TX
~ ..
A
PEL-
have the
UAP ; they
spacesc There are no known general
in other concrete spader The existence.
only for
verify the UAP
criteria which can be used to
{xi}ni=1
X-UAP for some X
p (~, )
these
method which works
a
X-UAP if there is
vectors
n
an
unconditional basis does not
The
are
only
(besides
spaces
known to have the UAP
compact
operators
disc algebra have
on
l2
ensure
those. cloare
refle-
with the usual
the UAP ?
closely related. This follows from the observation
has the UAP if and only if X** has the UAP (this is known to be
is replaced by AP)c In this connection, I would like to mention also.
X
false if UAP
are
V
(Q.9) Let
X
positive if
pass
ble spaces
X~
have
we
now
restrict ourselves
to
have
to
the UAP ? It is known that the
uniformly convex spaces ~2 a e
topic (d) - existence of bases. Of
failing
to have
the AP
implies
course
answer
is
the existence of separa-
in
particular that there are separable
basis. A quite simple fact which was known
already to
Banach is that every infinite-dimensional Banach
space has a subspace with a basis.
spaces which fail to have
This fact is
one
a
of the very few known results which
"infinite-dimensionel objects"
which
general Banach space (there are
the existence of concrete finite-dimensional
objects in
ensure
spaces,
e.
g.
the
(Q.10) Let
on
existence of
following question
X
a
Dvoretzky’s theorem) Q
the result
ticular,
in
the existence of nice
ensure
an
results
general Banach
investigate
be improved. In par-
It is therefore of much interest to
a
subspace.
with
a
basis
can
arises.
be an infinite-dimensional Banach
dimensional subspace with
more
space.
unconditional basis.?
Does
X
have
an
infinite-
A recent result of MAUREY and ROSENTHAL
may be
showed that
negative. They
[13] suggests
natural
a
that the
(Q.10)
to
answer
approach to constructing unconditional
basic sequences does not work in general (they exhibited a sequence of elements
= 1 for every n and
in a Banach space so
lim; x = 0 but
n n=1
subsequence
’
no
A
{xnk}~k=1
question closely
X
Let
be
an
that ~xn
~~ n ,r~
{xn }~n=1 forms
of
(Q.1.0)
related to
Y
following.
either reflexive
that
so
An old result of JAMES shows that
(Q.11).
to
answer
has
a
deepest
positive
or
one
contain
X
infinite-
an
isomorphic to one of the
A different
is
a
question
to
coordinate system
lent to
a
if
recall that
series 03A3n 03BBn
= 1
related to
(a
are C0, l1 and l2
for every
n ).
If
a
i.
e.
while
(Q.11).
in what
Banach space
a
is said to ba
with {03BBn}~n=1
xn
only if 03A3n x n yn converges.. Unconditional bases
unique. In fact, the only Banach spaces which have up
normalized if
are
uniqueness,
besis {xn}~n=1
a
and
lized unconditional basis
answer
negative
a
generated by a suitable basis in
me
basis
has
positive
who characterized those Banaeh spaoes which
[15]
bases is concerned with their
on
a
~~ .
nical coordinate system. Let
a
(Q.10)
(Q.10) implies
several results which
are
is due to ROSENTHAL
to
answer
that
unlikely
There
answer.
positive
a
It is however not
subspace isomorphic
a
sense
is the
sequence).
Co and
spaces
have
unconditional basic
infinite-dimensional Banach space. Does
dimensional subspace
The
n
an
a
cano-
equiva-
scalars converges if
if
to
they exist are rarely
equivalence a unique norma-
basis
Banach space
{xn }~n=1
X
is. said to be
has two
non
equiva-
lent normalized unconditional bases then it is
easily seen to have a normalized
unconditional basis
which is not symmetric (an unconditional basis
is said to be symmetric if
is equivalent to
for every permutation n of the integers). It is then easy to see that the family
where n ranges over all permutations of the integers contains uncountably many
mutually non equivalent normalized unconditional bases on X . If we restrict ourselves to
There
are
symmetric bases then
examples of
many
the phenomenon of
spaces which:have up
uniqueness is much more common.
to equivalence a unique symmetric
basis and thus in them the symmetric basis provides a canonical coordinate
system;
(this is the case e. go
Lorentz
sequence spaces, many Orliex
v p; ~ ~ ?
In
some.
sequence spaces).
situations (even for
Orlicz sequence spaces), a space.
may have more than one symmetric basise In all known examples of
more
spaces
than
one
symmetric
basis there
tric Bases. Unlike the situation
we
could
to be
simply
use
uncountably many mutually
are
with
permutations to
non
having
equivalent
uniqueness of unconditional bases ( were:
generate, new equivalence classes) ~ there seems
non
general procedure to generate new symmetric bases if we know that there
exist two non equivalent ones. This .lives rise to the
following question.
no
X
be
a
Banach space having
at least
two
non
equivalent symm
etric bas .
Must
X
(For
have infinitely many mutually non equivalent symmetric bases ?
further’
We pass
take
as
There-
to
now
subsets
convex
background
to
(Q.’!2),
cf.
[’H]).
’
linear problems. The structure of a Banach space and its
metric spaces has been a subject of much research (of course, we
some non
as
the metric the natural
induced
one
by
d(x, y)
direction (cf.
the norm, i.
many strong and beautiful theorems in this
are now
[2~ Actually,
the study of Banach spaces
as
e.
metric spaces created
=
t)x -
the book
a new
branch if
topology called infinite-dimensional topology. That this subject is a branch- of topology rather than Banach space theory stems from. the fact that the linear topological.
properties of a Banach space have almost no topological meaning. For example, I
recall the result of KADEC which states that any two separable, infinite-dimensional
Banach spaces are homeomorphic. The situation changes
completely if we take. into
account not
the
only
topology induced by
but the uniform structure indu--.
)j x it.
The constructions used in infinite dimensional
by
topology are almost never
uniformly continuous. It is not true that every two infinite-dimensional. separable
ced
Banach spaces
space
uniformly homeomorphic.
are
gives much information
it determines it
(Q.13) Let
on
its
fact, the uniform structure of a Banach
linear structure, and it is not known whether
completely*
X
Y
In
.
be two uniformly homeomorphic Banach.
phic to Y ?
spaces. Is.
X
isomor-
The strongest known partial result to
(Q.13) is a result of RIBE [14] which says
that if X is uniformly homeomorphic to Y then the local
structure of X and Y
as Banach spaces are the same
(i. e.. there is a constant X m so that for every
subspace B of X with dim B eo ther; is a B’ c y with d(B, B’) B ~. An
interesting special
spaces have the
case).
both satisfy
AHARONI
case
same
[t]
of
Lipschitz
this). Moreover
(it non linear subset)
uniformly homeomorphic to
matter of
spaces. Let
fact,
and
V=C(0 , t) (these
so
(it
is well known that there is no linear T
there is a Lipschitz continuous projection from X
TV . These results seem. to suggest that
may be
C
*!) .
there is very little known
mention just
me
= C
T"~
onto
a
X
RIBE’s result does not say anything in this
proved that there is mapping T : Y X so that T and
condition
a
is obtained if
local structure
which, satisfies
As
(Q.13)
one more
(among
on
the uniform structure of Banach
the very many
natural) question
on
this
subject.
(Q.14) Is the Hilbert space l2 uniformly homeomorphic
’
self ?
AHARONI
[1]
observed that the
We come. to cur last
answer is
positive if
topic-non separable questions.
attracted much less attention than separable
ones
we
Non
to
a
bounded subset of
"201420142014’""201420142014201420142014201420142014
replace
~ by C .
separable questions. have
among the research, in Banach space,
the fact
knowledge. on non separable; spaces is illustrated by
that the following simple looking problem is still, apparently open.
The lack of
theory.
(Q.1~)
X
Does every infinitely dimensional- Banach space
have
an
infinite. dimensio-
naL separable quotient space ?
In many
tive and
special case.s (,e;e g. if X is reflexive),
trivially so but no general construction of
the
(,Q.1;5)
to
answer
is
posiis
separable quotient space
a
known.
extensively studied class of non separable Banach space for which a
kind of structure theory is available is the class of WCG. spaces (spaces which.
have a weakly compact set which generates the whole spacer This class includes all.
The most
reflexive
L1( )
g.
is
where
(Q.16)
X
topology 3 is
TALAGRAND
finite
a
WCG
of
subspace
In the
X
isomorphic to
showed
(e;.
other spaces
on an
subspace of
a
is
X
is Lindelöf in its
Lindelbf space in its
a
w
WCG space ?
a
recently that conversely
space)
WCG
a
as
WCG
every
w
space
(ard
thus every
topology.
separable case., the question of classifying up to. isomprphism. the spaseems to pose very difficult questions. In: the separable case
(i. e.
non
C(K)
ces
a
as well
arbitrary measure space). Among the:
spaces which are still open’I mention the following.:
measure
Banach space. Assume that
be
[117]
w - compact)
the unit ball is
problems concerning
many
K
(where
spaces
metric),
compact
a
complete classification
up to
BESSAGA and PELCZYNSKI and MLLUTIN. The main step
K
C.(p )
is the Cantor set.
where
p
metrizabla)
Hausdorff
non
g. if
is the space of ordinals
K
was
was
carrie.d out
by
the result of MILUTIN which
compact metric uncountable then C(K)
showed that if
is
isomorphism
is
isomorphic to
In the non-separable case ( i. e. K compact
K (e.
some information on special classes of
in the order topology, or a topological. group or
there is
.
compactification of a metric space or an Eberlein compact). The general classification problem f.or non separable C(,K). spaces seems to be hopelessly
complicated. However there are some concrete questions which may have a reasonable
answer. The key point in the proof of MILUTIN’s result was that every separable
C(,K.). spaoe is isoporphia to such a space with K totally disconnected. Is this
true in general ?
a
Stone-Cech
(Q.17)
Let
K
be
a
ted
compact Hausdorf
In the setting of
existence of
(Q.18)
Does there exist..
compac t Hausdorff
f spa ce
non
K
separable
"nice
spaces, there
We mention here
Characterize those Banach
spaces
totally disconnec-
C(K) ~ C(K0) ?
tha t
so
a
one
are
many open
questions
about the
question of this type.
which have
an
equivalent strictly
))x
+
convex
norm.
A
norm
~Q.1~8).
parable
is
is of
stictly
convex
course a
Banach space
if
=
=
1~
and
y~I
=. 2
imply
that
x
=
y .
vaguely stated problem. It is easily verified that every sehas an equivalent strictly convex norm. The same is true for
18-09
general
a
shown
was
strictly
WCG
by
a
also for duals of
spaces);.
WCG
On the other
DAY that there exist Banach spaces which, do not have
Some conjectures
convex norm.
shown to be false in
be
(and
space
[3]o
concerning
This paper shows that
even
delicate and
there exists
an
possible
a
presumably difficult question to
equivalent strictly convex norm.
for
an
answer
C(K)
b.and, it
equivalent
to.
(Q.18)
spaces it
were
seame
to
deciae under. which condition
BIBLIOGRAPHY
[1]
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(I.). - Every separable
[2]
metric space is
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p
spaces and their applications, Studia Math., t. 29, 1968, p. 275-326.
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une
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(Texte
Joram LINDENSTRAUSS
Mathematics Department
Hebrew University
JERSUSALEM (Israel)
regu le 29 avril
1976)